How do I verify the trigonometric identity?

Verify the trigonometric identity:
secx(cscx-2sinx)=cotx-tanx

4 Answers
Apr 1, 2018

Using the identities:
1/cosx= secx
1/sinx= cscx
1-sin^2x= cos^2x
sinx/cosx= tanx
cosx/sinx= cotx

Start:
secx(cscx-2sinx)= cotx-tanx

1/cosx(1/sinx-2sinx)=cotx-tanx

1/(sinxcosx)-(2sinx)/cosx=cotx-tanx

1/(sinxcosx)-(2sin^2x)/(cosxsinx)=cotx-tanx

(1-2sin^2x)/(cosxsinx)=cotx-tanx

(1-sin^2x-sin^2x)/(cosxsinx)=cotx-tanx

(cos^2x-sin^2x)/(cosxsinx)=cotx-tanx

(cos^2x)/(cosxsinx)-(sin^2x)/(cosxsinx)=cotx-tanx

cosx/sinx-sinx/cosx=cotx-tanx

cotx-tanx= cotx-tanx

Apr 2, 2018

See below.

Explanation:

When verifying such identities, it's best to leave one side alone, and begin working with the more complex side. In this case, that would be the left side.

So, recall that secx=1/cosx, cscx=1/sinx:

1/cosx(1/sinx-2sinx)=cotx-tanx

Distribute:

1/(cosxsinx)-(2sinx)/cosx=cotx-tanx

Subtract the two expressions on the left side, using cosxsinx as a common denominator.
(1-2sin^2x)/(cosxsinx)=cotx-tanx

Recall the identity cos2x=1-2sin^2x=cos^2x-sin^2x. Apply to the numerator:

(cos^2x-sin^2x)/(cosxsinx)=cotx-tanx

Now, split up the fraction:

cos^2x/(cosxsinx)-sin^2x/(cosxsinx)=cotx-tanx

Simplify by cancelling out relevant terms:

cosx/sinx-sinx/cosx=cotx-tanx

Recalling that cotx=cosx/sinx, tanx=sinx/cosx, this holds:

cotx-tanx=cotx-tanx

Apr 2, 2018

Use the definition of the trig functions to rewrite the problem:

secx=1/cosx

cscx=1/sinx

tanx=sinx/cosx

cotx=1/tanx=1/(sinx/cosx)=cosx/sinx

and also the Pythagorean identity:

cos^2+sin^2x=1

=>1-sin^2x=cos^2x

Now, rewrite the problem in terms of sine and cosine. I'll start with the left side and manipulate it until it looks exactly like the right side:

LHS=color(red)secx(color(blue)cscx-2sinx)

color(white)(LHS)=color(red)(1/cosx)(color(blue)(1/sinx)-2sinx)

color(white)(LHS)=color(red)(1/cosx)*color(blue)(1/sinx)-color(red)(1/cosx)*2sinx

color(white)(LHS)=1/(cosxsinx)-color(red)(1/cosx)*2sinx

color(white)(LHS)=1/(cosxsinx)-(2sinx)/cosx

color(white)(LHS)=1/(cosxsinx)-2*sinx/cosx

color(white)(LHS)=1/(cosxsinx)-2*tanx

color(white)(LHS)=1/(cosxsinx)-tanx-tanx

color(white)(LHS)=1/(cosxsinx)-sinx/cosx-tanx

color(white)(LHS)=1/(cosxsinx)-(sinxcolor(red)(*sinx))/(cosxcolor(red)(*sinx))-tanx

color(white)(LHS)=1/(cosxsinx)-sin^2x/(cosxsinx)-tanx

color(white)(LHS)=(1-sin^2x)/(cosxsinx)-tanx

color(white)(LHS)=cos^2x/(cosxsinx)-tanx

color(white)(LHS)=cos^color(red)cancelcolor(black)2x/(color(red)cancelcolor(black)cosxsinx)-tanx

color(white)(LHS)=cosx/sinx-tanx

color(white)(LHS)=cotx-tanx

color(white)(LHS)=RHS

The identity is proved. Hope this helped!

Apr 4, 2018

LHS=secx(csc-2sinx)

=1/cosx(1/sinx-2sinx)

=1/(cosxsinx)-(2sinx)/cosx

=(cos^2x+sin^2x)/(cosxsinx)-2tanx

=cos^2x/(cosxsinx)+sin^2x/(cosxsinx)-2tanx

=cosx/sinx+sinx/cosx-2tanx

=cotx+tanx-2tanx

=cotx-tanx=RHS